chemistry, please help
posted by apoorva .
2NO2(g)> 2NO(g) + O2(g) , H=+114.2kJ
(Note: H, S, G all have a degree sign next to them)
NO: H(enthalpy)=90.3kJ/mol, S(entropy)=210.7J/mol*K, G(gibbs energy)=86.6
O2: H(enthalpy)=0, S(entropy)=?, G(gibbs energy)=0 kJ/mol
NO2: H(enthalpy)=33.2, S=239.9, G=51
For part a, I have to calcuate G for the reaction.
I used the Gibbs free standard reaction formula:where i have to find the sum of products minus the sum of reactants.
I got 71.2 kJ/mol. I know I got that right because my friend got the same answer too.
for part b, I have to calculate standard entropy, S for only O2.
There's a table in my book saying that entropy of O2 is 205 J/mol*K.
But I have to solve this using formulas and I was wondering if you can help on how to approach this problem.

i forgot to write down another part of question 
apoorva
The temperature of the reaction is 298 kelvin

tried solving it, please check answer 
apoorva
This is what I did:
my teacher said that this is a two step calculation:
First step:equation of heat
q=mct
m=mass of O2=32 grams
T=298+273=573 Celsius
q=32 g of O2(4.184J/g C)(573 C)
q=76450J/mol
then I used the equation:
S(entropy)=q/T
S=76450J/mol / (298 K)
=257 J/mol*K
I looked at a table in my chemistry book, and it says that S of O2 is 205 J/mol*K.
I'm assuming that's the answer I should reach, but I didn't.
So please see what I did wrong.
Thank you 
chemistry, please help 
drwls
298 K is not 573 C. It is 25 C
Entropy at 298 requires an integration of C dT/T from 0 K to 298, including solid and liquid phases. It cannot be calculated with a simple formula. The 205 J/mol K number for 298 K agrees with my JANNAF Tables.
Respond to this Question
Similar Questions

Chem
I'm unbelievably stuck on this problem. Any help would be appreciated. Red phosphorous is formed by heating white phosphorous. Calculate the temperature at which the two forms are at equilibrium, given white P: (heat of formation)= … 
Chemistry
Calculate the entropy change in the surroundings when 1.00 mol N2O4(g) is formed from 2.00 mol NO2(g) under standard conditions at 298 K. I get +192 J/K. The book says 192 J/K. Here is my work, where am I going wrong? 
ap chem
calculate the lattice enthalpy of potassium fluoride from the following data: enthalpy of formation of K(g): +89 kJ · mol−1 first ionization energy of K(g): +418 kJ · mol−1 enthalpy of formation of F(g): +79 kJ · mol−1 … 
Chemistry
Ammonia (NH3) boils at 33oC; at this temperature it has a density of 0.81 g/cm3. The enthalpy of formation of NH3(g) is 46.2kj/mol, and the enthalpy of vaporization of NH3(l) is 23.2 kJ/mol. Calculate the enthalpy change when 1 L … 
chemistry:)
The bond enthalpy of the Br–Cl bond is equal to DH° for the reaction BrCl(g)> Br(g) + Cl(g). Use the following data to find the bond enthalpy of the Br–Cl bond. Br2(l)>Br2(g) ÄH=30.91 KJ/mol Br2(g)>2Br2(g) ÄH=192.9 … 
chemistry:)
The bond enthalpy of the Br–Cl bond is equal to DH° for the reaction BrCl(g)> Br(g) + Cl(g). Use the following data to find the bond enthalpy of the Br–Cl bond. Br2(l)>Br2(g) ÄH=30.91 KJ/mol Br2(g)>2Br2(g) ÄH=192.9 … 
Chemistry Problem
Consider an ionic compound, MX, composed of generic metal M and generic halogen X. The enthalpy of formation of MX is ÄHf° = –457 kJ/mol. The enthalpy of sublimation of M is ÄHsub = 121 kJ/mol. The ionization energy of M is IE … 
Chemistry
Calculate the standard entropy, ΔS°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn, is –44.2 kJ·mol–1. C2H4(G)+H20 > C5H5OH ΔS°rxn= … 
chemistry
A method for scrubbing CO2(g) from the air on a spacecraft is to allow CO2(g) to react with NaOH according to the following (unbalanced reaction). NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l) Use the appropriate thermodynamic tables … 
chemistry
The enthalpy of formation of MX is ΔHf° = –493 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 143 kJ/mol. The ionization energy of M is IE = 437 kJ/mol. The electron affinity of X is ΔHEA = –317 kJ/mol. (Refer …